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I'm trying to solve this trigonometric equation: $$\sin(12º)\sin(24º)\sin(84º-xº) = \sin(30º)\sin(30º)\sin(xº)$$

I got here after applying Trigonometric Ceva Theorem. Here, I don't know how to solve it, I tried to use $\sin(xº)\sin(60º-xº)\sin(60º+xº)=\frac{\sin(3xº)}{4}$, but I couldn't proceed further.

Any hints are appreciated.

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  • $\begingroup$ Do you have two $\sin 30^\circ$? Isn't it $\sin^230^\circ=1/4$? $\endgroup$ – Quang Hoang Dec 9 '19 at 17:35
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    $\begingroup$ Yes, but I put the complete equation after applying the Ceva theorem to give a little bit of context. $\endgroup$ – Rodrigo Pizarro Dec 9 '19 at 17:36
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    $\begingroup$ With such setups (Trig form of Ceva), you usually have to guess what $x$ is first, and then prove the trigonometric identity. It is often hard to stare and it and see something magically appear, esp working with sin (84-x). (We know a unique solution exists because of monotonicity of sin ( 84 - x) / sin x.) In this case, $ x = 18^ \circ$. $\endgroup$ – Calvin Lin Dec 9 '19 at 17:39
  • $\begingroup$ I knew the answer, I just want to know how to get there after guessing it. $\endgroup$ – Rodrigo Pizarro Dec 9 '19 at 17:42
  • $\begingroup$ In which case, update your question accordingly. Also, typically you just brute force your way through. The hard part is guessing the value. $\endgroup$ – Calvin Lin Dec 9 '19 at 17:51
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$$\dfrac{\sin x}{\sin(84-x)}=4\sin12\sin24$$

Using the Werner Formulas, $4\sin y\sin2y=2(\cos y-\cos3y)$

Using Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$

$$2\cos12-2\cos36=2\cos12-(1+2\cos72)=2\sin42-1$$

$$2\sin42=\dfrac{2\sin42\cos(42-x)}{\sin(84-x)}$$

I should leave it to you now!

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HINT: Once you get into the form $$\frac{\sin(x)}{\sin(84 - x)}$$ Use the trig identity $$\sin(A-B) = \sin(A)\cos(B) - \cos(A)\sin(B)$$ and then try to cancel the $\sin(x)$ in the numerator. I was able to simplify after this, although its a bit of a pain.

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